12 J
The work done by a constant force is calculated using the formula:
\(W = F \times d \times \cos(\theta)\)
Where:
- \(W\) is the work done (in Joules)
- \(F\) is the force applied (in Newtons)
- \(d\) is the displacement (in meters)
- \(\theta\) is the angle between the force and the displacement vectors.
In this problem, we are given:
- \(F = 6 \, N\)
- \(d = 2 \, m\)
- \(\theta = 0^\circ\) (since the displacement is in the direction of the force)
Substituting these values into the formula, we get:
\(W = 6 \, N \times 2 \, m \times \cos(0^\circ)\)
Since \(\cos(0^\circ) = 1\), the equation simplifies to:
\(W = 6 \, N \times 2 \, m \times 1 = 12 \, J\)
Therefore, the work done is 12 Joules.
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Work done will be maximum when the angle between the direction of force and the direction of direction of displacement is
Important Questions from Work
A body of mass 100 kg rests on a horizontal plane, the value of coefficient of friction between the body and plane being 0.025. Find the work done in moving the body through a distance of 10 metres along the plane.
A load of 16.5 kg is lifted through a height of 3.4 metres. Find the work done in kg metre.
What is defined as the rate of doing work or the rate of transfer of energy?
______ efforts are where our eyes direct the movement of our bodies.
Task simplifications represent _______ work methods.
